Question

A particle of specific charge (qm) is projected from the origin of coordinate with initial velocity $[u\hat{i}-v\hat{j}]$ Uniform electric magnetic fields exist in the region along the +y direction, of magnitude E and B. The particle will definitely return to the origin once if.

• A. $\frac{BE}{2E}$ is an integer
• B. $(u^2+v^2{}^{1/2})\left[B/E\right]$
• C. $\left[VB/E\right]$ in an integer
• D. $\left[uB/Eisaninteger\right]$

Answer 1

The correct option is A $\frac{BE}{2E}$ is an integer

Charge $=q_m$

Velocity $=u\hat{i}-u\hat{j}$

along the $+y$ direction $E$ and $B$.

The particle will definitely return to the origin once if.

In $Y$ direction the electric field $E$ exist. So, the force due to electric on the charge is given by

$F=qE$

Now, the acceleration of the charge is given by $a=\frac{qE}{m}$

now, the acceleration on is opposite to the direction of velocity. So, it will return to origin if displacement in $y-$ direction is zero.

$d=V_y\times t+\frac{1}{2}{at}^2$

$0=-V\times t+\frac{1}{2}\frac{qE}{m}{T}^2$

$T=\frac{2mV}{qE}$

now due to magnitude field it will move in circular path, time period of its circular motion is given by

$T=\frac{2\pi m}{qB}$

now if complete $N$ number of rounds in above time of return then

$N\times \frac{2\pi m}{qB}=\frac{2mV}{qE}$

$N=\frac{Bv}{\pi E}$

So, here above value must be an integer so that it complete integral number of rounds

integer $=\frac{Bv}{\pi E}=\frac{Bv}{2E}$ ($\because$ $n=2$) in an integer.